If $$a$$ and $$b$$ are the roots of the equation $$Px^2 - Qx + R = 0$$, then what is the value of $$\left(\frac{1}{a^2}\right) + \left(\frac{1}{b^2}\right) + \left(\frac{a}{b}\right) + \left(\frac{b}{a}\right)$$?

According to question :

$$a+b=\frac{Q}{P}\ and\ ab=\frac{R}{P}\ .$$

So, $$\left(a^2+b^2\right)=\frac{Q^2}{P^2}-\frac{2R}{P}=\frac{\left(Q^2-2PR\right)}{P^2}\ .$$

And, $$\left(\frac{1}{a^2b^2}+\frac{1}{ab}\right)=\left(\frac{P^2}{R^2}+\frac{P}{R}\right)=\frac{P\left(P+R\right)}{R^2}\ .$$

So, $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{b}{a}+\frac{a}{b}$$

$$=\left(\frac{b^2+a^2}{a^2b^2}\right)+\left(\frac{b^2+a^2}{ab}\right)$$

$$=\left(a^2+b^2\right)\left(\frac{1}{a^2b^2}+\frac{1}{ab}\right).$$

$$=\frac{\left(Q^2-2PR\right)}{P^2}.\ \frac{P\left(P+R\right)}{R^2}$$

$$=\frac{\left(Q^2-2PR\right)\left(P+R\right)}{PR^2}\ .$$

B is correct choice.